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[I know it's is a Physics Problem but I am sure that I am committing mistake in integration part. That's why so posting here. I am new to calculus-based physics so I frequently make conceptual mistakes when setting up integrals; I would really appreciate someone pointing them out.]

Goal: Finding the center of mass of a semicircular wire/disk of on non-negligible width, with the inner radius being $R_1$ and outer radius being $R_2$.

  1. I am gonna start this with a goal of setting up a Reimann Sum. First I divide the "arc" of angle $\pi$ into $n$ sub-arcs of equal angle $\Delta\theta$

  2. The total center of mass can be found if centers of mass of parts of the system are known. In each circular arc interval, I choose a height, $H_i$, approximating the height of the center of mass of each sub arc, hoping that the error goes to $0$ in the limit as $n \to \infty$, and multiply this by the mass of the sub arc. Pushing this through the limiting process, I set up the integral of $H$ w.r.t $m$.

  3. Finding $H_i$: Now, as $\Delta\theta \to0$, the sector-difference region formed by each sub-arc should get closer and closer to a tilted rectangle. Assuming that to be $\color{green}{\text{True}}$, the center of mass of each sub-arc (being approximated by a titled rectangle) would be a distance $H_i=\frac{(R_1+R_2)\sin(θ)}{2}$ above the origin.

  4. Lastly, since the shape has a constant mass per unit area, the differential mass and total mass can be replaced by differential area and total area. Using the sector area formula for each subinterval, the differential area, $dA$, should be equal to $0.5\;d\theta (R_2^2-R_1^2)$

Solving this gives me $$y_{com}=\frac{(R1+R2)}{\pi}$$ which upon looking up is clearly wrong. It is interesting thought that it gives the correct result when $R_1=R_2$ ($0$-thickness). What is the error in my reasoning?

Here I have solved it on Pen and Paper.
enter image description here enter image description here

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3 Answers 3

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For sufficiently small $\Delta\theta,$ your element of area is approximately an isosceles trapezoid with one parallel side $R_1\Delta\theta$ and opposite side $R_2\Delta\theta$. No matter how small $\Delta\theta$ gets, the larger parallel side will always be $R_2/R_1$ times as large as the other.

If the parallel sides of a trapezoid have lengths $a$ and $b,$ and the interior of the trapezoid is a lamina of uniform density, then the center of mass of that lamina is at a distance $$ \frac{2a + b}{3(a+b)} h$$ from the side of length $b,$ where $h$ is the height of the trapezoid. Let $a = R_2\Delta\theta$ and $b = R_1\Delta\theta$, these sides being respectively at distances $R_2$ and $R_1$ from the center of the circle; then $a > b,$ $h = R_2 - R_1,$ and the center of mass is at a distance

$$ \frac{2R_2\Delta\theta + R_1\Delta\theta} {3(R_2\Delta\theta+R_1\Delta\theta)} (R_2 - R_1) =\frac{2R_2+ R_1}{3(R_2+R_1)}(R_2 - R_1) $$

from the short side of the trapezoid, which is is the same for every small $\Delta\theta$ and is greater than $\frac12(R_2 - R_1).$ Hence the center of mass is located farther than $\frac12(R_2 + R_1)$ from the center of the circle.

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Due to symmetry, the center of mass lies along the radial line of the angle $\frac{\Delta\theta}2$. So, it is natural to let the $y$-axis aligned with the half-angle radius. Then, integrate the center of mass in polar coordinates as follows,

$$y_c = \frac{\int_{R_1}^{R_2}\int_{\frac{\pi-\Delta\theta}2}^{ \frac{\pi+\Delta\theta}2}( r\sin\theta )rdr d\theta}{\frac{\Delta \theta}2(R_2^2-R_1^2)}=\frac43\frac{}{}\frac{\sin\frac{\Delta\theta}2}{\Delta\theta}\frac{R_2^3-R_1^3}{R_2^2-R_1^2} $$

(You incorrectly assumed that $\frac{R_2+R_1}2$ is the COM along the radial direction, because the mass within $d\theta$ is not uniform along radius.)

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  • $\begingroup$ yes, i agree that the mass isnt uniform. But shouldnt it become uniform in the limit as Δθ goes to 0? I know this is the entire idea behind the definite integral, but knowing when the error goes to 0 (like rectangles approximating area) and when it doesnt(like rectangles approximating arc length) gets me every time. $\endgroup$ Commented Jul 25, 2020 at 17:24
  • $\begingroup$ @OVERWOOTCH - the error does not go to zero in limit. $\endgroup$
    – Quanto
    Commented Jul 25, 2020 at 17:28
  • $\begingroup$ wait, are we talking about the same Δθ? your final answer includes it as a variable which makes me believe you assumed it be angle of the total arc. I have kept the angle pi radians and taken Δθ to be the angle of each interval which gets infinitely small in the limit. pls correct me if im wrong $\endgroup$ Commented Jul 25, 2020 at 17:31
  • $\begingroup$ @OVERWOOTCH - the result given in the answer Is valid for arbitrary $\Delta\theta$. You may set it to $pi$, or let it go to zero limit. $\endgroup$
    – Quanto
    Commented Jul 25, 2020 at 17:34
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    $\begingroup$ @OVERWOOTCH For sufficiently small $\Delta\theta,$ your element of area is approximately an isoceles trapezoid with one parallel side $R_1\Delta\theta$ and opposite side $R_2\Delta\theta.$ No matter how small $\theta$ gets, the larger parallel side will always be $R_2/R_1$ times as large as the other, and the COM will always be closer to the larger end of the trapezoid, never anywhere near the halfway point between the two ends. $\endgroup$
    – David K
    Commented Jul 25, 2020 at 18:40
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Let us start with the center of mass of a unit half-disk. By symmetry, the average abscissa is $0$. Then the average ordinate,

$$\bar y=\frac1A\int_{-1}^1\int_0^{\sqrt{1-x^2}}y\,dx\,dy=\frac2\pi\frac12\int_{-1}^1(1-x^2)\,dx=\frac4{3\pi}.$$

To obtain the result for a half ring of radii $R_e$ and $R_i$, we will combine a positive and a negative mass. Computing the moments and areas, this gives us

$$\bar y=\frac{M_{y,e}-M_{y,i}}{A_e-A_i}=\frac4{3\pi}\frac{R_e^3-R_i^3}{R_e^2-R_i^2}=\frac4{3\pi}\frac{R_e^2+R_eR_i+R_i^2}{R_e+R_i}.$$

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  • $\begingroup$ thankyou for your answer, but I am aware of different ways of finding it. I am more interested in singling out the error in my reasoning so it doesnt happen again $\endgroup$ Commented Jul 25, 2020 at 18:30

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